Predicting the Products of Electrolysis

What Is Electrolysis?

In order to predict the products of electrolysis, we first need to understand what electrolysis is and how it works. Electrolysis is a method of separating bonded elements and compounds by passing an electric current through them. It uses a direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction. Electrolysis is very important commercially as a stage in the separation of elements from naturally occurring sources, such as ores, using an electrolytic cell.

The main components required to achieve electrolysis are:

• An electrolyte: a substance containing free ions, which are the carriers of electric current in the electrolyte. If the ions are not mobile, as in a solid salt, then electrolysis cannot occur.
• A direct current (DC) supply: provides the energy necessary to create or discharge the ions in the electrolyte. Electric current is carried by electrons in the external circuit.
• Two electrodes: an electrical conductor that provides the physical interface between the electrical circuit providing the energy and the electrolyte.

The Interchange of Atoms and Ions

The key process of electrolysis is the interchange of atoms and ions by the removal or addition of electrons to the external circuit. The required products of electrolysis are in a different physical state from the electrolyte and can be removed by some physical processes.

Each electrode attracts ions that are of the opposite charge. Positively charged ions, or cations, move toward the electron-providing cathode, which is negative; negatively charged ions, or anions, move toward the positive anode. You may have noticed that this is the opposite of a galvanic cell, where the anode is negative and the cathode is positive.

At the electrodes, electrons are absorbed or released by the atoms and ions. Those atoms that gain or lose electrons become charged ions that pass into the electrolyte. Those ions that gain or lose electrons to become uncharged atoms separate from the electrolyte. The formation of uncharged atoms from ions is called discharging. The energy required to cause the ions to migrate to the electrodes, and the energy to cause the change in ionic state, is provided by the external source.

Oxidation and Reduction

Oxidation of ions or neutral molecules occurs at the anode, and reduction of ions or neutral molecules occurs at the cathode. For example, it is possible to oxidize ferrous ions to ferric ions at the anode:

$Fe^{2+} (aq) \rightarrow Fe^{3+} (aq) + e^-$

It is also possible to reduce ferricyanide ions to ferrocyanide ions at the cathode:

$Fe(CN)^{3-}_6 + e^- \rightarrow Fe(CN)^{4-}_6$

Neutral molecules can also react at either electrode. Electrolysis reactions involving H⁺ ions are fairly common in acidic solutions. In alkaline water solutions, reactions involving hydroxide ions (OH^-) are common. The substances oxidized or reduced can also be the solvent, which is usually water, or the electrodes. It is possible to have electrolysis involving gases.

Predicting the Products of Electrolysis

Let's look at how to predict the products. For example, what two ions will CuSO₄ break down into? The answer is Cu²⁺ and SO₄^2-. Let's look more closely at this reaction.

We take two copper electrodes and place them into a solution of blue copper sulfate (CuSO₄) and then turn the current on. We notice that the the initial blue color of the solution remains unchanged, but it appears that copper has been deposited on one of the electrodes but dissolved on the other. This is because Cu²⁺ ions are attracted to the negatively charged cathode, and since the the cathode is putting out electrons, the Cu²⁺ becomes reduced to form copper metal, which is deposited on the electrode. The reaction at this electrode is:

$Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$

Electrolysis of copper sulfate

Two copper electrodes are placed in a solution of blue copper sulfate and are connected to a source of electrical current. The current is turned on for a period of time.

At the positive anode, copper metal is oxidized to form Cu²⁺ ions. This is why it appears that the copper has dissolved from the electrode. The reaction at this electrode is:

$Cu (s) \rightarrow Cu^{2+} (aq) + 2e^-$

We just saw electric current used to split CuSO₄ into its component ions. This is all it takes to predict the products of electrolysis; all you have to do is break down a compound into its component ions.